From the equation [tex] A \vec{V} = \lambda \vec{V} [/tex], where [tex] \vec{V} = (x, y) [/tex] I get :

[tex] -3x - 5y = -1x + i \sqrt{11}x [/tex]

[tex] 3x + y = -1y + i \sqrt{11}y [/tex]

Which gives:

[tex] -2x - i \sqrt{11}x - 5y = 0[/tex]

[tex] 3x + 2y - i \sqrt{11}y = 0 [/tex]

When I solve this system for x and y, I get a solution of (0, 0). The book agrees with the eigenvalue that I found, but has an eigenvector solution of [tex] (-2 + i \sqrt{11}, 3) [/tex]. Can anyone spot what I'm doing wrong?

If I solve the first equation for y, I get
[tex]y = -\frac{1}{5} (2 + \mathrm{i}\sqrt{11}) x [/tex]
Then plugging this into the second equation gives an equation just for x. Don't forget to simplify the prefactor as much as possible, then solve for x. You'll see that though (x, y) = (0, 0) is a possibility (which you don't want, because you want it to be an eigenvector), but there are also others for which x is non-zero.

Also note, that once you found one eigenvector, you can take any multiple and it will be an eigenvector again. So you can multiply the whole thing by a factor to make the vector look nicer (e.g. if you'd get [itex](12/\sqrt{1 + x}, \sqrt{1 - x})[/itex] I'd multiply by [itex]\sqrt{1 + x}[/itex] and write it as [itex](12, \sqrt{x^2 + 1})[/itex]).

Thanks for the replies. I realize what I was doing wrong (to some degree). I made an algebraic mistake along the way. My answer wasn't the same answer that the book had, but it worked such that [tex] A \vec{V} = \lambda \vec{V} [/tex], which makes me assume that my answer was a multiple of the books answer.